On the Positive Solutions of the Free-Boundary Problem for Emden-Fowler Type Equations

Yi Li^{∗}, Hans G. Kaper, and Man Kam Kwong^{†}

1 Introduction

Let Ω be a smooth, bounded and connected domain in ℜ^{n}. In this paper, we consider the following boundary value problem:

\{ \begin{matrix}Δu + f(u) = 0 & in Ω, \\ u > 0 & in Ω, \\ u = \frac{∂u}{∂n} = 0 & on ∂Ω. \end{matrix}  de1  (1.1)

Here, n denotes the unit outer normal to ∂Ω. (See [KK] for existence and uniqueness results for (1.1).) We prove the following two theorems.

Theorem 1 Let f be such that

f(s) = f_{1}(s) + f_{2}(s),  f  (1.2)

where f_{1} is nondecreasing and f_{2} Lipschitz continuous. If u ∈ C^{2}(\overline{Ω}) be a classical solution of (1.1), then Ω is an open ball, Ω = B_{R}(x_{0}) say, in ℜ^{n} and u is radially symmetric about the center x_{0}. Furthermore,

\frac{∂u}{∂r} < 0 for 0 < r ≡ |x − x_{0}| < R.

Theorem 2 Let B_{R}(0) be a ball of radius R > 0. Let u be a classical solution of the boundary value problem,

\{ \begin{matrix}Δu + f(u) = 0 & in B_{R}(0), \\ u > 0 & in B_{R}(0), \\ u = 0 & on ∂B_{R}(0). \end{matrix}  de2  (1.3)

∗Research supported in part by the National Science Foundation

†This work was supported by the Office of Scientific Computing, U.S. Department of Energy, under Contract W-31-109-Eng-38.

If f ∈ C_{loc}^{0,1}((0, ∞)) and there exists an s_{0} > 0 such that f(s) is strictly decreasing in [0, s_{0}], then u is radially symmetric about 0. Furthermore,

\frac{∂u}{∂r} < 0 for 0 < r < R.

We use the well-known moving-plane method, which was first proposed by Alexandrov. In 1971, Serrin used this method to prove the symmetry result for (1.1) in the case where f(s) is real and constant. Since we are dealing with nonsmooth functions like f_{1}, some stronger version of the Hopf near-boundary theorem has to be used. In fact, we use the moving-plane method, in combination with a result (Lemma 4) of Gidas, Ni and Nirenberg [GNN], to prove Theorem 1. To prove Theorem 2, we need to analyze the locations of possible minima of the difference u − u^{λ} in order to continue the moving-plane process.

2 Preliminaries

To prove Theorems 1 and 2, we need a series of technical lemmas, whose proofs can be found in [GNN], [H], [PW] and [S].

Let u be a nonnegative classical solution in Ω of the following differential inequality:

Lu ≡ a^{ij}(x)D_{ij}u + b^{i}(x)D_{i}u + c(x)u ≤ 0 in Ω,  di  (2.1)

where (a^{ij}(x)) ≥ λI in Ω for some fixed λ > 0 and a^{ij}, b^{i}, c ∈ L^{∞}(Ω).

Lemma 1 (Strong Maximum Principle) If u ≥ 0 in Ω and u vanishes at some point inside Ω, then u ≡ 0 in Ω.

Lemma 2 (Hopf Boundary Lemma) If x_{0} ∈ ∂Ω, u > 0 in Ω, and u(x_{0}) = 0, then

\frac{∂u}{∂n}(x_{0}) < 0

and

lim_{x→x_{0},x∈Ω} \frac{u(x) − u(x_{0})}{|x − x_{0}|} > 0  (2.2)

for any non-tangential limit.

Lemma 3 ([GNN]) Let Ω be a domain in ℜ^{n}. Let y_{0} ∈ ∂Ω and assume that, near y_{0}, ∂Ω consists of two transversally intersecting C^{2}−hypersurfaces φ = 0 and ψ = 0. Suppose that φ, ψ < 0 in Ω. Let u satisfy (2.1), u > 0 in Ω, and u(y_{0}) = 0. Assume that

ℓ(y_{0}) ≡ a^{ij}(y_{0})D_{i}φ(y_{0})D_{j}ψ(y_{0}) ≥ 0

and, if ℓ(y_{0}) = 0, assume furthermore that a^{ij} ∈ C^{2} in some Ω_{ε} = Ω ∩ B_{ε}(y_{0}) (ε > 0), and that ∇_{t}(ℓ(y)) = 0 at y_{0} for any tangential derivatives ∇_{t} along {φ = 0} ∩ {ψ = 0}. Then

\frac{∂u}{∂s} > 0 at y_{0} if ℓ(y_{0}) > 0,

\frac{∂u}{∂s} > 0 or \frac{∂^{2}u}{∂s^{2}} > 0 at y_{0} if ℓ(y_{0}) = 0,

for any direction s entering Ω at y_{0} transversally to the hypersurfaces φ = 0 and ψ = 0.

Lemma 4 ([GNN]) Let x_{0} ∈ ∂Ω with n_{1}(x_{0}) > 0, where n_{1} is the first component of n. Assume that u > 0 in Ω_{ε}, u ≡ 0 on ∂Ω ∩ B_{ε}(x_{0}), and

Δu + f(u) = 0 in Ω,

where f satisfies (1.2). Then there exists a δ > 0 such that D_{1}u < 0 in Ω_{δ}.

3 Proof of Theorem 1

Before we can use the moving-plane method, we introduce a few definitions. Let e_{1} = (1, 0, …, 0) be the unit vector along the x_{1}-axis, and let T_{λ} be the hyperplane {x_{1} = λ}. Since Ω is bounded and smooth, T_{λ} ∩\overline{Ω} = ∅ for large λ. Now, let λ decrease until T_{λ} touches \overline{Ω} at λ_{0}, say.

For λ < λ_{0}, let Σ_{λ}^{+} = Ω ∩ {x_{1} > λ}, and let Σ_{λ}^{−} be the reflection of Σ_{λ}^{+} about the plane T_{λ}. Let x^{λ} be the reflection point of x about T_{λ}, i.e.,

x^{λ} = (2λ − x_{1}, x_{2}, …, x_{n}).

If λ_{0} − λ is small, Σ_{λ}^{−} will be inside Ω. But as λ decreases continuously, Σ_{λ}^{−} will be in Ω until one of the following occurs:

1. \overline{Σ_{λ}^{−}} becomes internally tangent to ∂Ω at some point x_{0} not on T_{λ};

2. T_{λ} becomes orthogonal to ∂Ω at some point y_{0} ∈ ∂Ω ∩ T_{λ}.

We let T_{λ_{1}} denote the plane that first reaches one of these two possibilities and call Σ_{λ_{1}}^{+} the maximal cap.

Lemma 5 If

u(x) ≤ u(x^{λ}) in Σ_{λ}^{+}

for some λ ∈ (λ_{1}, λ_{0}), then

\{ \begin{matrix}D_{e_{1}}u(x) < 0 & on Ω ∩ T_{λ}, \\ u(x) < u(x^{λ}) & in Σ_{λ}^{+}. \end{matrix}  iq2  (3.1)

Proof. Let v(x) = u(x^{λ}), x ∈ Σ_{λ}^{+}. Then

Δ(v(x) − u(x)) + f_{1}(v(x)) + f_{2}(v(x)) − f_{1}(u(x)) − f_{2}(u(x)) = 0

and

v(x) − u(x) ≥ 0 in Σ_{λ}^{+},

with

v(x) − u(x) ≢ 0 in Σ_{λ}^{+}.

Now, f_{1}(v(x)) ≥ f_{1}(u(x)) and f_{2}(v(x)) − f_{2}(u(x)) = c_{λ}(x)(v(x) − u(x)), where c_{λ}(x) is bounded, because f_{1} is nondecreasing and f_{2} is Lipschitz. Hence,

Δ(v(x) − u(x)) + c_{λ}(x)(v(x) − u(x)) ≤ 0,  iq3  (3.2)
v(x) − u(x) ≥ 0 in Σ_{λ}^{+},  iq4  (3.3)

and

v(x) − u(x) ≢ 0 in Σ_{λ}^{+}.  (3.4)

Then Lemma 1 implies that v(x) − u(x) > 0 in Σ_{λ}^{+}, while Lemma 2 gives us that

\frac{∂}{∂x_{1}} (v(x) − u(x))|_{x_{1}=λ} > 0,  iq5  (3.5)

because v(x) − u(x)|_{x_{1}=λ} ≡ 0, which is the minimum. From (3.5), we obtain the inequality

−\frac{∂u}{∂x_{1}} − \frac{∂u}{∂x_{1}} |_{x_{1}=λ} > 0,

which completes the proof of Lemma 5.

Lemma 6 Let u be a classical solution of (1.1). Then (3.1) holds for all λ ∈ (λ_{1}, λ_{0}).

Proof. From Lemma 4, we know that (3.2) and (3.3) hold for all λ ∈ (λ_{1}, λ_{0}) with λ_{0} − λ sufficiently small.

Suppose the lemma is false. That is, suppose that there exists a λ_{2} ∈ (λ_{1}, λ_{0}) such that (3.1) holds for λ ∈ (λ_{2}, λ_{0}) but not for λ < λ_{2}.

On the other hand, the continuity of u implies that

u(x) ≤ u(x^{λ_{2}}), λ ∈ Σ_{λ_{2}}^{+},

and since λ_{2} ∈ (λ_{1}, λ_{0}), Lemma 5 implies that (3.2) and (3.3) also hold for λ = λ_{2}.

Since λ_{2} > λ_{1}, n_{1}(x_{0}) > 0 for each point x_{0} ∈ ∂Σ_{λ_{2}}^{+}\(T_{λ_{2}} ∩ Ω). And hence Lemma 4 concludes that there exists ε_{x_{0}} > 0 such that

D_{e_{1}}u(x) < 0 in Ω ∩ B_{ε_{x_{0}}} (x_{0}).

Since D_{e_{1}}u(x) < 0 on T_{λ_{2}} ∩ Ω, we have that there exists some ε > 0, such that

D_{e_{1}}u(x) < 0 in Ω ∩ {x_{1} > λ_{2} − ε},  iq6  (3.6)

because T_{λ_{2}} ∩ \overline{Ω} is compact.

Therefore, if (3.2) and (3.3) fail to hold in (λ_{1}, λ_{2}), we must have a sequence {λ^{i}} such that

λ^{i} > λ_{1} and λ^{i} ↗ λ_{2} as i → ∞,

with

u(x^{i}) ≥ u(x^{iλ^{i}}) for some x^{i} ∈ Σ_{λ_{i}}^{+}.

But Ω is bounded, so we can find a subsequence of {x^{i}}, say {x^{i}} itself, which converges to some point x_{0} ∈ \overline{Σ_{λ_{2}}^{+}} as i → ∞ with u(x_{0}) ≥ u(x_{0}^{λ_{2}}). Therefore, x_{0} ∈ ∂Σ_{λ_{2}}^{+}, because (3.1) holds for λ_{2}. Thus, we have either of two possibilities:

1. x_{0} ∈ ∂Σ_{λ_{2}}^{+}\(T_{λ} ∩ \overline{Ω}). But then x_{0} ∈ ∂Ω with x_{0}^{λ_{2}} ∈ Ω, since λ_{2} > λ_{1}, which implies that 0 ≥ u(x_{0}^{λ_{2}}) > 0. This is impossible.

2. x_{0} ∈ T_{λ_{2}} ∩ \overline{Ω}. Therefore, x_{0}^{λ_{2}} = x_{0}.

Since λ^{i} > λ_{1}, we have that the line segment P_{i} joining x^{i} and x^{iλ^{i}} lies in Ω. Therefore, u(x^{i}) ≥ u(x^{iλ^{i}}) implies that

D_{e_{1}}u(y^{i}) ≥ 0 for some y^{i} ∈ P_{i}.

But x^{i} → x_{0} and x^{iλ^{i}} → x_{0}^{λ_{2}} = x_{0}, so P_{i} shrinks into the single point x_{0}. This gives us a contradiction with (3.6), because y^{i} ∈ Ω ∩ {x_{1} > λ_{2} − ε} for i large enough. This completes the proof of Lemma 6.

Proof of Theorem 1. By Lemma 6, (3.2) and (3.3) hold for all λ ∈ (λ_{1}, λ_{0}). Let us discuss the following two possible cases.

Case 1. Σ_{λ_{1}}^{−} ∪ Σ_{λ_{1}}^{+} ∪ (T_{λ_{1}} ∩ Ω) = Ω.

Then Ω is symmetric about T_{λ_{1}}, in which case we have shown that

\{ \begin{matrix}u(x) ≤ u(x^{λ_{1}}), & x ∈ Σ_{λ_{1}}^{+}, \\ D_{e_{1}}u < 0, & x ∈ Σ_{λ}^{+}, \end{matrix} ∀ λ ∈ (λ_{1}, λ_{0}),  iq7  (3.7)

or

\{ \begin{matrix}u(x) = u(x^{λ_{1}}), & x ∈ Ω, \\ D_{e_{1}}u > 0, & if x ∈ Ω ∩ Σ_{λ_{1}}^{−}, \\ D_{e_{1}}u < 0, & if x ∈ Ω ∩ Σ_{λ_{1}}^{+}. \end{matrix}  iq8  (3.8)

Case 2. Σ_{λ_{1}}^{−} ∪ Σ_{λ_{1}}^{+} ∪ (T_{λ_{1}} ∩ Ω) ⊂_{≠} Ω.

Then u(x^{λ_{1}}) − u(x) ≥ 0 in Σ_{λ_{1}}^{+} and not identically zero. Therefore, the same argument as in Lemma 5 implies that

u(x^{λ_{1}}) > u(x) in Σ_{λ_{1}}^{+}.

at some point x_{0} ∉ T_{λ_{1}}. As in the proof of Lemma 5, we find by letting v(x) = u(x^{λ_{1}}) that

\{ \begin{matrix}Δ(v(x) − u(x)) + c_{λ_{1}}(x)(v(x) − u(x)) ≤ 0 & in Σ_{λ_{1}}^{+}, \\ v(x) − u(x) > 0 & in Σ_{λ_{1}}^{+}, \\ v(x_{0}) − u(x_{0}) = 0. \end{matrix}  iq9 (3.9)

Since x_{0} ∉ T_{λ_{1}}, Σ_{λ_{1}}^{+} is smooth near x_{0}. Hence we may use Lemma 2 to conclude that

\frac{∂}{∂n}(v − u)(x_{0}) < 0,

which is in contradiction with the boundary condition \frac{∂v}{∂n} = \frac{∂u}{∂n}(x_{0}) = 0. Hence, T_{λ_{1}} must become orthogonal to ∂Ω at some point y_{0} ∈ ∂Ω ∩ T_{λ_{1}}.

However, u(x^{λ_{1}}) − u(x) satisfies (3.8) in Σ_{λ_{1}}^{+}, y_{0} ∈ ∂Σ_{λ}^{+}, and, near y_{0}, ∂Σ_{λ}^{+} consists of two transversally intersecting hypersurfaces x_{1} = λ_{1} and ∂Ω, which become orthogonal at y_{0}. A simple computation shows that ℓ(y_{0}) = 0 and, for any tangential direction t along T_{λ_{1}} ∩ ∂Ω at y_{0},

∇_{t}(ℓ(y)) = 0 at y_{0},

which implies by Lemma 3 that for any s entering Ω at y_{0} transversally to T_{λ} and ∂Ω,

\frac{∂(v − u)}{∂s} > 0 or \frac{∂^{2}(v − u)}{∂s^{2}} > 0 at y_{0}.

On the other hand, it follows from (1.1) that

(v − u)(y_{0}) = 0, ∇(v − u)(y_{0}) = 0, and D^{2}(v − u)(y_{0}) = 0.

This again leads us to a contradiction, so it must be the case that

Σ_{λ_{1}}^{−} ∪ Σ_{λ_{1}}^{+} ∪ (T_{λ_{1}} ∩ Ω) = Ω.

On the other hand, since we can start moving the plane from the left to the right along the x_{1}-axis as well, we conclude that

\{ \begin{matrix}u(x) = u(x^{λ_{1}}) & x ∈ Ω, \\ D_{e_{1}}u > 0 & if x ∈ Ω ∩ Σ_{λ_{1}}^{−}, \\ D_{e_{1}}u < 0 & if x ∈ Ω ∩ Σ_{λ_{1}}^{+}. \end{matrix}  iq10  (3.10)

But equation (1.1) is rotationally invariant. Therefore Ω is symmetric in every direction. We thus find that Ω must be a ball, because it is connected. Then (3.10) gives the conclusions of Theorem 1.

4 Proof of Theorem 2

Gidas, Ni, and Nirenberg proved in [GNN] that the solutions of

\{ \begin{matrix}Δu + f(u) = 0 & in B_{R}(0), \\ u > 0 & in B_{R}(0), \\ u = 0 & on ∂B_{R}(0), \end{matrix}  de3  (4.1)

with

f(s) = f_{1}(s) + f_{2}(s),  f2  (4.2)

where f_{1} is Lipschitz continuous and f_{2} nondecreasing, must be radially symmetric about 0 and, furthermore, \frac{∂u}{∂r} < 0 for 0 < r < R. On the other hand, if a decomposition like (4.2) does not exist, in particular if f is not smooth, then it is an open problem whether positive solutions of (1.1) are radially symmetric. Actually, some examples given in [GNN, pp. 220] show that these cases could be very delicate.

In this part, we will try to treat a family of nonlinear terms f which are neither Lipschitz nor nondecreasing. Such situations arise, for example, in the study of free-boundary problems for Emden-Fowler type equations (see [KK]), where

f(u) = u^{1/p} − u^{1/q} with 1 ≤ p < q ≤ ∞.  f3  (4.3)

Remark 1. If q = ∞, then f(s) = s^{1/p} − 1 is an increasing function in s and therefore [GNN]’s result implies that u must be radially symmetric. Therefore, the difficult parts occur when 1 ≤ p < q < ∞. For such cases,

f(s) = \{ \begin{matrix}strictly decreasing in [0, (p/q)^{pq/(q−p)}] , \\ strictly increasing in [(p/q)^{pq/(q−p)} , ∞) . \end{matrix}

Remark 2. Recently, new symmetry results have been obtained in [GL] and [LV] for equations on nonsmooth domains.

Proof of Theorem 2. First we define

Λ = {λ ∈ (0, R)|u(x) < u(x^{λ}) if x ∈ Σ_{λ}^{+}} .

Because u|_{∂B_{R}} = 0, there exists a λ_{0} ∈ (0, R) such that

u(B_{R}(0)\B_{λ_{0}}(0)) ⊂ (0, s_{0}). incl (4.4)

\underline{Step 1.} (\frac{1}{2}(λ_{0} + R), R) ∈ Λ.

For any λ ∈ (\frac{1}{2}(λ_{0} + R), R), f is a strictly decreasing function in teh interval [ 0, max { sup_{Σ_{λ}^{+}} u, sup_{S_{λ}^{+}} u^{λ} }], where u^{λ} = u(x^{λ}), because of (4.4), and

Δ(u^{λ} − u)(x) + f(u^{λ}(x)) − f(u(x)) = 0 in Σ_{λ}^{+},
u^{λ} − u = 0  on \overline{T}_{λ},
u^{λ} − u > 0  on ∂Σ_{λ}^{+} \ \overline{T}_{λ}.

\underline{Claim 1.} If u^{λ} − u ≥ 0 in Σ_{λ}^{+}, then u^{λ} − u > 0 in Σ_{λ}^{+} and \frac{∂u}{∂x_{1}} < 0 on T_{λ}.

Suppose the claim is false, i.e. there exists a y_{0} ∈ Σ_{λ}^{+}, such that (u^{λ} − u)(y_{0}) = 0.

On the other hand, both u and u^{λ} are strictly positive in Σ_{λ}^{+}, so

Δ(u^{λ} − u)(x) + \frac{f(u^{λ}(x)) − f(u(x))}{u^{λ}(x) − u(x)}(u^{λ} − u)(x) = 0,

where \frac{f(u^{λ}(x))−f(u(x))}{u^{λ}(x)−u(x)} is locally bounded, because f ∈ C_{loc}^{0,1}((0, ∞)).

Hence the strong maximum principle implies a contradiction. Therefore, if u^{λ} − u ≥ 0 in Σ_{λ}^{+}, then u^{λ} − u > 0 there.

\underline{Claim 2.} u^{λ} − u ≥ 0 in Σ_{λ}^{+}.

Otherwise, because u^{λ} − u ≥ 0 on ∂Σ_{λ}^{+}, u^{λ} − u would have a strictly interior negative minimum, say at y_{0} ∈ Σ_{λ}^{+}. But at y_{0} we have Δ(u^{λ} − u)(y_{0}) ≥ 0 and, since s_{0} > u(y_{0}) > u^{λ}(y_{0}) > 0 by (4.4), f(u(y_{0})) < f(u^{λ}(y_{0})). Therefore,

Δ(u^{λ} − u)(y_{0}) + f(u^{λ}(y_{0})) − f(u(y_{0})) > 0,

a contradiction.

Thus Step 1 is proved.

\underline{Step 2.} Λ is closed w.r.t. (0, R).

If {λ^{i}} is a sequence in Λ which converges to some λ in (0, R), then, since

u(x) < u(x^{λi}), x ∈ Σ_{λ_{i}}^{+},

letting i → ∞, we find

u(x) ≤ u(x^{λ}), x ∈ Σ_{λ}^{+}.

But then ‘Claim 1’ in ‘Step 1’ shows that u(x) < u(x^{λ}) in Σ_{λ}^{+}, i.e. λ ∈ Λ.

\underline{Step 3.} Λ is open in (0, R).

Suppose that Λ is not open. Then there exists a λ ∈ Λ and a sequence {λ^{i}} ∈ (0, R) s.t. λ^{i} → λ with λ^{i} ∉ Λ. That is, for each i there exists x^{i} ∈ Σ_{λ^{i}}^{+} with

0 > u(x^{iλ^{i}}) − u(x^{i}) = min_{x∈Σ_{λ^{i}}^{+}}(u^{λ^{i}} − u)(x),  min  (4.5)

and the mean-value theorem implies

\frac{∂u}{∂x_{1}} (y^{i}) ≥ 0 for some y^{i} ∈ \overline{x^{iλ^{i}}x^{i}}.  mvt  (4.6)

Because x^{i} ∈ B_{R}(0), there exists a subsequence, say {x^{i}} itself, converging to a point x_{0} ∈ \overline{B}_{R}(0) and (4.5) implies that

u(x_{0}) ≥ u(x_{0}^{λ}). iq11 (4.7)

But λ ∈ Λ, therefore (4.7) could only occur on ∂Σ_{λ}^{+}. Hence x_{0} ∈ ∂Σ_{λ}^{+}.

On the other hand,

u < u^{λ} on ∂Σ_{λ}^{+}\\overline{T}_{λ}.

Therefore x_{0} ∈ \overline{T}_{λ}.

In this case, since x^{i} → x_{0} ∈ \overline{T}_{λ} and λ^{i} → λ, we have x^{iλ^{i}} → x_{0}. Therefore (4.6) implies that

\frac{∂u}{∂x_{1}} (x_{0}) ≥ 0.

Hence, x_{0} ∈ \overline{T}_{λ} ∩ ∂B_{R}(0) , because

\frac{∂u}{∂x_{1}} < 0 in \overline{T}_{λ}.

Now, since x_{0} ∈ ∂B_{R}(0) and lim_{i→∞} x^{i} = lim_{i→∞} x^{iλ^{i}} = x_{0}, we have 0 < u(x^{iλ^{i}}) < u(x^{i}) < s_{0} if i is large enough. Therefore, f(u(x^{i})) < f(u(x^{iλ^{i}})) for all i large enough.

But Δ(u^{λ^{i}} − u)(x^{i}) ≥ 0, since x^{i} is a minimum point of u^{λ^{i}} − u, so we reach a contradiction, because

Δ(u^{λ} − u)(x^{i}) + f(u(x^{iλ^{i}})) − f(u(x^{i})) = 0.

Therefore Λ is open.

\underline{Step 4.} Since Λ is non-empty and both open and closed in (0, R), it must be the case that Λ = (0, R), so letting λ → 0, we find that

u(x_{1}, ⋯, x_{n}) ≤ u(−x_{1}, ⋯, x_{n})

for x ∈ Σ_{0}^{+}. But both B_{R}(0) and Δ are invariant under the symmetry group, so

u radially symmetric

and

\frac{∂u}{∂r} < 0 for 0 < r < R,

by ‘Claim 1’. Thus, the proof of Theorem 2 is complete.

References

[GL] N. Garofalo and J. Lewis, A Symmetry Result Related to Some Overdetermined Boundary Value Problems, Amer. J. Math. 111 (1989), 9–33.

[GNN] B. Gidas, W.-M. Ni and L. Nirenberg, Symmetry and Related Properties via the Maximum Principle, Comm. Math. Phys. 68 (1979), 209–243.

[H] H. Hopf, Lectures on Differential Geometry in the Large, Stanford Univ. (1956).

[KK] H. G. Kaper and Man Kam Kwong, Free Boundary Problems for Emden-Fowler Equations, Differential and Integral Equations 3 (1990), 353–362.

[LV] J. Lewis and A. Vogel, On Some Almost Everywhere Symmetry Theorems, preprint.

[PW] M. Protter and H. Weinberger, Maximum Principles in Differential Equations, Prentice-Hall (1967).

[S] J. Serrin, A Symmetry Problem in Potential Theory, Arch. Rational Mech. Anal. 43, (1971), 304–318.